If `f(x)={{:(2x^(2)+1",",xle1),(4x^(3)-1",",xgt1):}`, then `int_(0)^(2)f(x)dx` is

Options

A. `47//3`
B. `50//3`
C. `1//3`
D. `47//2`

Answer

To solve the problem, we need to evaluate the definite integral of the piecewise function \( f(x) \) from 0 to 2. The function \( f(x) \) is defined as:

\[ f(x) = \begin{cases} 2x^2 + 1 & \text{if } x \leq 1 \\ 4x^3 - 1 & \text{if } x > 1 \end{cases} \]

We will break the integral into two parts based on the definition of the function:

\[ \int_0^2 f(x) \, dx = \int_0^1 f(x) \, dx + \int_1^2 f(x) \, dx \]

### Step 1: Evaluate \( \int_0^1 f(x) \, dx \)

For \( x \) in the interval \([0, 1]\), we have:

\[ f(x) = 2x^2 + 1 \]

Thus, we can write:

\[ \int_0^1 f(x) \, dx = \int_0^1 (2x^2 + 1) \, dx \]

### Step 2: Integrate \( 2x^2 + 1 \)

Now, we calculate the integral:

\[ \int (2x^2 + 1) \, dx = \int 2x^2 \, dx + \int 1 \, dx \]

Calculating each part:

1. \( \int 2x^2 \, dx = \frac{2}{3} x^3 \)
2. \( \int 1 \, dx = x \)

Putting it together:

\[ \int (2x^2 + 1) \, dx = \frac{2}{3} x^3 + x \]

### Step 3: Evaluate from 0 to 1

Now we evaluate from 0 to 1:

\[ \left[ \frac{2}{3} x^3 + x \right]_0^1 = \left( \frac{2}{3} (1)^3 + (1) \right) - \left( \frac{2}{3} (0)^3 + (0) \right) \]

Calculating this gives:

\[ = \left( \frac{2}{3} + 1 \right) - 0 = \frac{2}{3} + \frac{3}{3} = \frac{5}{3} \]

### Step 4: Evaluate \( \int_1^2 f(x) \, dx \)

For \( x \) in the interval \([1, 2]\), we have:

\[ f(x) = 4x^3 - 1 \]

Thus, we can write:

\[ \int_1^2 f(x) \, dx = \int_1^2 (4x^3 - 1) \, dx \]

### Step 5: Integrate \( 4x^3 - 1 \)

Now, we calculate the integral:

\[ \int (4x^3 - 1) \, dx = \int 4x^3 \, dx - \int 1 \, dx \]

Calculating each part:

1. \( \int 4x^3 \, dx = x^4 \)
2. \( \int 1 \, dx = x \)

Putting it together:

\[ \int (4x^3 - 1) \, dx = x^4 - x \]

### Step 6: Evaluate from 1 to 2

Now we evaluate from 1 to 2:

\[ \left[ x^4 - x \right]_1^2 = \left( (2)^4 - (2) \right) - \left( (1)^4 - (1) \right) \]

Calculating this gives:

\[ = (16 - 2) - (1 - 1) = 14 - 0 = 14 \]

### Step 7: Combine the results

Now we combine the results from both integrals:

\[ \int_0^2 f(x) \, dx = \int_0^1 f(x) \, dx + \int_1^2 f(x) \, dx = \frac{5}{3} + 14 \]

To add these, we convert 14 into a fraction:

\[ 14 = \frac{42}{3} \]

Now we can add:

\[ \int_0^2 f(x) \, dx = \frac{5}{3} + \frac{42}{3} = \frac{47}{3} \]

### Final Answer

Thus, the value of the integral \( \int_0^2 f(x) \, dx \) is:

\[ \boxed{\frac{47}{3}} \]